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Math 181 Miniproject 3: Texting Lesson.md
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My lesson Topic
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<div><img class="left"/><div class="alert gray">

Hello tutor! I was just wondering if you could help me with this new concept we learned in my Calculus class yesterday I think I understand it, but I want to make sure.

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<div><div class="alert blue">

Hello student! I would be more than happy to help you. What is it that you learned?

</div><img class="right"/></div>

<div><img class="left"/><div class="alert gray">

I believe it was about Linear Approximation. I understand how we find the equation of the tangent line, but I don't understand how we use it to etsimate the value of more difficult problems that would otherwise need a calculator.

</div></div> 

<div><div class="alert blue">

Ok... first tell me what you do know and understand.

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<div><img class="left"/><div class="alert gray">

I know how to find the equation for a tangent line using the formula:
$y=f(a)+f'(a)(x-a)$
And when given a function I know how to find $L(x)$. But I don't understand how the Linear Approximation formula is used to find values for "ugly" numbers.

</div></div> 

<div><div class="alert blue">

Ok. So, using the linear approximation of a function to estimate more difficult values of that function comes from looking at a graph. If we can identify a point on the tangent line that is really close to that equivalent x-value on the function, than that y-value on the tangent would be pretty close to that y-value on the function.

</div><img class="right"/></div>

<div><img class="left"/><div class="alert gray">

Oh... I see. Could  you give me an example?

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<div><div class="alert blue">

Sure. Let's say we have the function $f\left(x\right)=\sqrt[3]{x}$ and its derivative is $f'\left(x\right)=\frac{1}{3\sqrt[3]{x^2}}$ at $x=1000$. What is the Linear Approximation?

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<div><img class="left"/><div class="alert gray">

$L(x)=f(a)+f'(a)(x-a)$
$L(x)=\sqrt[3]{1000}+\left(\frac{1}{3\sqrt[3]{1000^2}}\right)(x-1000)$
$L(x)=10+\frac{1}{300}x-1000$

I think...

</div></div>

<div><div class="alert blue">

Yes! Now if I told you to give me the y-coordinate, $f(x)=\sqrt[3]{1003}$, it would be extremely difficult to do without a calculator. However, using the Linear Approximation of x, we can estimate a coordinate on the tangent line of the function that is really close to $\sqrt[3]{1003}$.

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<div><img class="left"/><div class="alert gray">

Mmm...

</div></div>

<div><div class="alert blue">

So, being that $f(x)=\sqrt[3]{1003}$ we know x, in the instance is equal to $1003$. We then plug that into our equation

$L(1003)=10+\frac{1}{300}(1003-1000)$

Then solve.
$L(1003)=10.01$ therefore, $f\left(1003\right)\sim10.01$

</div><img class="right"/></div>

<div><img class="left"/><div class="alert gray">

Ohhhhh...that makes sense. You're not calculating the exact answer, you're just estimating based on the tangent since graphically it is so close.

</div></div>

<div><div class="alert blue">

Exactly. Good job

</div><img class="right"/></div>

<div><img class="left"/><div class="alert gray">

Thanks for helping me understand.

</div></div>

<div><div class="alert blue">

No problem!

</div><img class="right"/></div>


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